At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 24 knots and ship B is sailing north at 25 knots. How fast (in knots) is the distance between the ships changing at 5 PM?
I have tried multiple times but keep getting confused.
at t = 0
xA = -20 , dxA/dt = -24
yA = 0, dyA/dt = 0
xB = 0, dxB/dt=0
yB = 0 , dyB/dt = 25
Dx = differnce in x location
= 20 + 24 t
Dy = difference in y location
= 25 t
distance = (Dx^2+Dy^2)^.5
distance = [ 400+480t+576t^2+ 625t^2 ]^.5
= [1201 t^2 + 480 t + 400 ]^.5
d distance/dt = .5[2402t - 480]/distance
if t = 3
d distance/dt = just put in 3 for t
To solve this problem, we can use the concept of the rate of change. Specifically, we can use the derivatives to find the rate at which the distance between the two ships is changing.
Let's start by visualizing the situation at noon.
Ship A is 20 nautical miles due west of ship B. This means that at noon, the distance between the two ships is 20 nautical miles. We can represent this distance as D.
Since ship A is sailing west at 24 knots, we can represent its rate of change as dx/dt = -24 knots (negative sign because it is moving west).
Ship B is sailing north at 25 knots, so we can represent its rate of change as dy/dt = 25 knots.
To find the rate at which the distance between the two ships is changing (dd/dt), we need to find the derivative of the distance D with respect to time t.
dd/dt = sqrt[(dx/dt)^2 + (dy/dt)^2] <- This is the derivative of distance with respect to time.
Now we have all the ingredients to find the rate at which the distance is changing at 5 PM. To do this, we need to substitute the values of dx/dt and dy/dt into the equation:
dd/dt = sqrt[(-24)^2 + (25)^2] knots
dd/dt = sqrt[576 + 625] knots
dd/dt = sqrt[1201] knots
By calculating the square root of 1201, we find:
dd/dt ≈ 34.64 knots
Therefore, the distance between the two ships is changing at a rate of approximately 34.64 knots.